CHAPTER 45 - 10 - For a mass equal to that of the Sun, we...

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Unformatted text preview: For a mass equal to that of the Sun, we get 51. If we have only N = M/m neutrons, the Fermi energy is n 2 En = 3 EF N = 3 N h 3N 5 5 8mn πV 2/ 3 = 3N h 2 18N 2 π2 160m n R 2/ 3 . The gravitational energy will be the same, so the total energy is E = En + Eg = 3N h 2 18N 2 2 160m n R π 2/ 3 2 – 3G M . 5R We find the equilibrium radius from dE/dR = 0: 2 d E = – 3N h 18N dR 80m n R 3 π 2 2/ 3 2 N h2 9N + 3G M = 0, which gives R = 2 5R 16m n G M 2 π 2 2/ 3 . In terms of the mass, this is For a mass of 1.5 solar masses, we get R= 16 1. 67 × 10 kg ( 6.67 × 10 4 = 1.1 × 10 m = 11 km . – 27 8/ 3 6.63 × 10 – 34 J ∙ s – 11 2 2 30 N ∙ m 2/ kg ) (1. 5)(2. 0 × 10 kg ) 1/ 3 18 2/ 3 π2 p y d x = c d t ²θ R r θ F = GMm / r2 x 52. (a) A photon of energy E will have a mass m = E/c , and a momentum p = E/c = mc. We assume the deflection is small. Using the coordinate system shown in the figure, this means y ˜ R and r ˜ x + R . We want the deflection in the y­direction. When the photon is at the position x, the vertical component of the gravitational force will produce an impulse F dt in the y 2 2 2 2 time dt = dx/c. Thus we have dp = – (GMm/r ) cos θ dt y 2 ...
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CHAPTER 45 - 10 - For a mass equal to that of the Sun, we...

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